Limits preserve non-strict inequalities

Statement

Let $(a_n)$ and $(b_n)$ be real sequences that converge to real numbers $a$ and $b$. Assume that there is a natural number $N$ such that $a_n \le b_n$ for all $n \ge N$. Then $a \le b$.

Remarks

Strict inequalities do not pass to the limit. For example, $0 < 1/n$ for all $n \ge 1$, but $\lim 1/n = 0$. With a constant sequence $b_n = c$, it says: if $a_n \le c$ for large $n$, then $\lim a_n \le c$. The proof is by contradiction. If $a > b$, take $\varepsilon = (a - b)/2$. Then for large $n$ we get $b_n < a_n$.

Explanations

L1 human

If one sequence is always below another, its limit cannot be above the limit of the other. The limits can be equal, even when the terms are never equal.